Question

На нейтрализацию 80 мл 0.1н раствора щелочи потребовалось 5 мл раствора HCl.Определить молярность раствора HCl.

Answer 1

80 : 5 = x :0,1
х=1,6
Cн(NaOH)=1,6моль/л; Cм(NaOH)=1,6моль/л

Answer 2

Calculation of Molarity of HCl Solution

To determine the molarity of the HCl solution, we can use the concept of neutralization. Neutralization is a chemical reaction between an acid and a base, resulting in the formation of a salt and water. In this case, the acid is HCl and the base is the 0.1N solution of alkali.

Given: - Volume of the alkali solution (NaOH) = 80 mL - Volume of the HCl solution = 5 mL

To find the molarity of the HCl solution, we can use the formula:

Molarity (M) = (moles of solute) / (volume of solution in liters)

First, let's calculate the moles of NaOH using the given volume and molarity:

Moles of NaOH = Molarity (N) * Volume (L)

Since the volume is given in mL, we need to convert it to liters by dividing by 1000:

Moles of NaOH = 0.1N * (80 mL / 1000 mL/L)

Now, let's calculate the moles of HCl using the volume and molarity:

Moles of HCl = Molarity (N) * Volume (L)

Again, we need to convert the volume to liters:

Moles of HCl = Molarity (N) * (5 mL / 1000 mL/L)

Since the reaction between HCl and NaOH is 1:1, the moles of HCl and NaOH will be equal.

Therefore, the molarity of the HCl solution is 0.1N.

Please note that the search results did not provide a direct answer to the question. However, the calculation of molarity is a basic concept in chemistry and can be solved using the given information and the formula mentioned above.

Answer 3

Calculation of Molarity of HCl Solution

To determine the molarity of the HCl solution, we can use the concept of neutralization. Neutralization is a chemical reaction between an acid and a base, resulting in the formation of a salt and water. In this case, the acid is HCl and the base is the 0.1N solution of alkali.

According to the given information, 80 mL of 0.1N alkali solution required 5 mL of HCl solution for neutralization.

To calculate the molarity of the HCl solution, we can use the following formula:

M1V1 = M2V2

Where: - M1 is the molarity of the alkali solution - V1 is the volume of the alkali solution - M2 is the molarity of the HCl solution - V2 is the volume of the HCl solution

Substituting the given values into the formula, we have:

0.1N * 80 mL = M2 * 5 mL

Simplifying the equation, we get:

8 = 5M2

Dividing both sides by 5, we find:

M2 = 8/5

Therefore, the molarity of the HCl solution is 1.6N.

Please note that the sources provided did not directly answer the question. The calculation was performed using the given information and the concept of neutralization.